OFFSET
0,2
COMMENTS
Binomial transform of A081567.
Case k=3 of family of recurrences a(n) = (2k+1)*a(n-1) - A028387(k-1)*a(n-2) for n >= 2, with a(0) = 1 and a(1) = k + 1.
a(n) = 4^n*a(n;1/4) = Sum_{k=0..n} binomial(n,k) * (-1)^k * F(k-1) * 4^(n-k), which also implies Deléham's formula given below and where a(n;d), n = 0, 1, ..., d, denote the delta-Fibonacci numbers defined in comments to A000045 (see also Witula's et al. papers). - Roman Witula, Jul 12 2012
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..200
Edyta Hetmaniok, Bożena Piątek, and Roman Wituła, Binomials Transformation Formulae of Scaled Fibonacci Numbers, Open Mathematics, 15(1) (2017), 477-485.
Roman Witula and Damian Slota, delta-Fibonacci numbers, Appl. Anal. Discr. Math 3 (2009), 310-329, MR2555042.
Index entries for linear recurrences with constant coefficients, signature (7,-11).
FORMULA
a(n) = 7*a(n-1) - 11*a(n-2) for n >= 2, with a(0) = 1 and a(1) = 4.
a(n) = (1/2 - sqrt(5)/10)*(7/2 - sqrt(5)/2)^n + (sqrt(5)/10 + 1/2)*(sqrt(5)/2 + 7/2)^n = A099453(n) - 3*A099453(n-1).
G.f.: (1 - 3*x)/(1 - 7*x + 11*x^2).
a(n) = Sum_{k=0..n} A094441(n,k)*3^k. - Philippe Deléham, Dec 14 2009
G.f.: Q(0,u)/x - 1/x, where u = x/(1 - 3*x), Q(k,u) = 1 + u^2 + (k+2)*u - u*(k + 1 + u)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 07 2013
E.g.f.: exp(7*x/2)*(5*cosh(sqrt(5)*x/2) + sqrt(5)*sinh(sqrt(5)*x/2))/5. - Stefano Spezia, Jun 03 2024
MAPLE
seq(coeff(series((1-3*x)/(1-7*x+11*x^2), x, n+1), x, n), n = 0 .. 30); # G. C. Greubel, Aug 12 2019
MATHEMATICA
CoefficientList[Series[(1-3x)/(1 -7x +11x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Aug 09 2013 *)
LinearRecurrence[{7, -11}, {1, 4}, 30] (* Harvey P. Dale, Feb 01 2015 *)
PROG
(Magma) I:=[1, 4]; [n le 2 select I[n] else 7*Self(n-1)-11*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Aug 09 2013
(PARI) Vec((1-3*x)/(1-7*x+11*x^2) + O(x^30)) \\ Altug Alkan, Dec 10 2015
(Sage)
def A081568_list(prec):
P.<x> = PowerSeriesRing(ZZ, prec)
return P((1-3*x)/(1-7*x+11*x^2)).list()
A081568_list(30) # G. C. Greubel, Aug 12 2019
(GAP) a:=[1, 4];; for n in [3..30] do a[n]:=7*a[n-1]-11*a[n-2]; od; a; # G. C. Greubel, Aug 12 2019
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Mar 22 2003
STATUS
approved